Solve the following :
Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.
Solution
To obtain the points of intersection of the line and the parabola, we equate the values of x from both the equations.
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1). Required area:area of the region OCABO
= area of the region OCADO – area of the region OBADO
Now, area of the region OCADO
= area under the parabola y2 = x i.e. y = `± sqrt(x)` (in the first quadrant) between x = 0 and x = 1
= `int_0^1 sqrt(x)*dx`
= `[(x^(3/2))/(3/2)]_0^1`
= `(2)/(3) xx (1 - 0)`
= `(2)/(3)`
Area of the region OBADO
= area under the line y = x between x 0 and x = 1
= `int_0^1x*dx`
=`[x^2/2]_0^1`
=`(1)/(2) - 0`
= `(2)/(3)`
∴ required area = `(2)/(3) - (1)/(2)`
= `(1)/(6)"sq unit"`.